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MATHEMATICS Proceedings A 84 (3), September 28, 1981
About graded fields
by J. van Gee1 and F. van Oystaeyen
University of Antwerp, 2630 Wilrijk, Belgirrm
Communicated by Prof. J.P. Murre at the meeting of September 27, 1980
INTRODUCTION In this paper we developwhat may be called “graded commutative algebra”. Our techniqueswill be different from the existing ones becausewe stay within the graded context i.e. we set up a theory about graded rings starting from the study of graded fields. The aim of this is the study of the (graded) Brauer group of a graded field and its relation to the Brauer groups of the quotient field and of the field of degree 0 in the graded field. Further results on these Brauer groups will be in a forthcoming paper, here in Section4 we give an introduction to this theory, including a graded version of Posner’s theorem for graded P.I. rings. The theory of algebraic graded extensions of graded fields given in Section 1 is useful in finding graded splitting fields of graded simple graded Artinian algebras.We have included graded versionsof Zariski’s theorem and Hilbert’s nulstellensatz;let us point out however that the graded nulstellensatz included here is different from the projective version of the theorem commonly used in projective geometry. Indeed, the affine “graded” geometry over graded fields is not a projective geometry over a common field (it can be shown that the affine geometry over a graded field may be considered as an hyperplane in Proj R for a certain graded ring R).
Although-we do not go into the geometric theory here we do include our results on graded valuation rings in graded fields, but here mainly becauseof their usefulnessin determining integral closuresand graded integral closures.
273 Another advantage of the concept of working consequently in the graded context is that one may obtain short proofs for some known theorems with otherwise tedious proofs. For example, the well-known (and very useful in projective geometry) theorem stating that: the integral closure of a graded domain in its field of fractions is again a graded domain (cf. Theorem 11, [ 141) turns out to be a specialcase of a corollary to our results on integral closure and graded integral closure. The theory of graded valuation rings, cf. Section 3, yields an application of the theory of primes in rings expoundedby the authors in [7], [S] and in [lo], since a graded valuation ring in a graded field is a prime of that graded field. We assumeelementary facts about graded rings and refer to [S] for details and further references. In this paper all graded rings are Z-graded!
1. GRADED FIELDS All rings will be associative with units. Unless otherwise stated we will considercommutative rings. A graded ring R is saidto be a graded division ring if every homogeneous element of R is invertible. A commutative graded division ring is called a graded field. The structure of a graded division ring is given in Theorem 4.1.) in the commutative casehowever it is easy to establish that a graded field K is either of the form Ko[t, t- ‘3 whereKO is a field, the part of degree0 in K, and a variable given positive degree, or elseK = Ko. Let R be a graded field, let K be its field of fractions and let 52be an algebraic closure of K. An element a~ $2 is said to be graded algebraic over R if the minimal manic polynomial of a over K has the form: with aiER and deg (ai)-deg (ai+l)=cEH, for all i=O,...,n-1. The latter condition states exactly that f(X) is a homogeneouselement of R[X] equipped with the gradation RIXln = C i.+jcFn RiXj, n E H. A ring S containing R is said to be graded algebraic over R if S is graded and everys E S is graded algebraicover R. In thesedefinitions it may at first sight be restricting to demand that exactly the minimal manic polynomial over K is of the prescribedtype instead of merely demandingthe existenceof such a manic polynomial. Apart from the fact that this will not work, the following lemma justifies the definition we have chosen;with notations as above:
LEMMA 1.1. If g(X) is a manic polynomial in R[X] with minimal degreein X such that g(a) = 0, then g(X) is a minimal polynomial for a over K.
PROOF. If g(X) factorizes as hl(X)hz(X) in R[X], then the leading co- efficients of hl and hz are units in R, hencethey haveto be homogeneous.Since R is a graded field we may assumethat hi(X) and hz(X) are manic polynomials, but then the minimality assumption on degx g(X) yields that either hi or h2 equals1, i.e. g(X) is irreducible in R[X]. Now R = Ro[ T, T-l] is a factorial ring and therefore g(X) will still be irreducible in K[X], cf. [4] p. 127. Consequently g(X) is a minimal polynomial for a over K. I 274 COROLLARY 1.2. An a E 52is graded algebraic over R if it is integral over R and the manic polynomial of minimal degree in X satisfied by a has the form (*).
COROLLARY 1.3. If arz Q is graded algebraic over R then the ring R[a] (constructed within 9) is a graded field.
PROOF. Let f(X), the minimal manic polynomial over R satisfied by a, be of the form (*). By the above lemma we have that R[a] zR[X]/(f(X)) as a ring. Moreover, if we equip R[X] with the gradation: R[X], = C i+jc=n RiXj, n E Z, then (j(X)) is a graded ideal of R[X] and this makes R[a] into a graded ring with deg (a) = C. If we establishthat cf(X)) is maximal as a graded ideal of R[X] then R[a] is a graded field. Therefore, consider a homogeneoush(X) ER[X]. The lemma entails that K[X](h(X), f(X)) equals either K[X] or K[X]df(X)). In the first casethe graded ideal u(X)), h(X)) containsa homogeneouselement of R i.e. an unit of R, thus in this case(h(X), f(X)) = R[X]. In the secondcase there exists an r ER such that r/r(X) E cf(X)), consequently:r&(X) E (j(X)) for each homo- geneouscomponent ri of r. SinceR is a graded field h(X) E (j(X)) follows. Thus (j(X)) is a maximal graded ideal of R[X]. H
LEMMA I .4. Let a E Q be graded algebraicover R and of degreec,, let PE 52 be graded algebraic over R and of degreecp, then p is graded algebraic over R[a] and of degreeCD.
PROOF. Let h be a manic polynomial for j?over R[a] and supposethat h has minimal degree in X as such. Since R[a] is a graded field (Corollary 1.3.), R[a][X] is a factorial ring and thus h is irreducible (Lemma I. 1.) and also h is the minimal polynomial for fl over K(a) (Lemma 1.1. again), where K(a) denotes the field of fractions of R[a]. Let gf R[X] be the minimal manic polynomial for j? over R, then g is homogeneousin R[X] if the latter ring is graded by putting R[X],, = Ci+jcszn RiXi. We embed R[X] into R[a][X] in the obvious way. In R[a][X] we have then that g = h< + p with p(J) = 0, hencep = 0 follows by the minimality condition on degx h. The ring R[a] [X] is graded such that deg (X) = cg, thus g = ht implies that h and { are homogeneousin R[a][X] i.e. h has the desiredform (A) over R[a]. Consequentlyp is graded algebraic over R[a] of degreeC,T. n
COROLLARY 1.5. The set of graded algebraicelements over R within 52is a graded field.
PROOF. Put E(R)c = (a E 0, a graded algebraic over R, deg (a) = c}. Take a,/? E E(R)c and considerR[a, p] in a. From the foregoing results it follows that R[a,j?] is a graded field, thus cr+p is homogeneousof degreec in R[a,/?]. If C r=-, ri(o + /?)’ is the minimal manic polynomial for a + p over R then by
275 selectingterms of highest degree,N say, we obtain: j+,lN %(a +P)‘=O where ri,j is the homogeneouscomponent of degreej of r;. Up to multiplication by a unit of R we may assumethat f(X) = Cj+ic=N QjX’is a manic polynomial of the form (*) which is satisfied by o+ #I, hence a+ DE JY(R)~. In much the same way it may be established that, (xEEL and beE( yields C@EE(R),+d. Finally, it follows that E(R) = @cE~E(R)c is a graded ring, evena graded field. n
REMARK 1.6. E(R) is termed to be the graded algebraic closure of R in 52. The mere existenceof a graded algebraic closure could have been established more directly, as in the ungraded case.
LEMMA 1.7. Every graded module over a graded field is graded free i.e. has a basisconsisting of homogeneouselements.
PROOF. Easy. H
As a consequenceof the above lemma we may talk about the (graded) rank of a graded module M over a graded field, we will denote it by [A&R]. The following statementsare easily verified: 1. If the degreein X of the minimal polynomialf(X) for somegraded algebraic a~ 52equals n then n = [R[cl]:R]. 2. If ai , . . .. an is a finite number of graded algebraic elements of Q then [R[cr~,.a., cxn]:R] < 00. 3. If S is a graded field extension of R such that [SRI < 03 then S is a graded algebraicextension of R. Now let R be a non-trivially graded field, Kits field of fractions and L a finite field extensionof K with [L:K”J= n. Put EL(R) = 0 CEz(~%(R))c, where (EL(R))~ = (czEL, IY is graded algebraic over R of degreec). Then EL(R) is a graded field and we refer to it as the graded algebraicclosure of R within L. With these notations:
PROPOSITION 1.8. [&(R):R] 5 [L:KJ = n.
PROOF. Straightforward. m
A field extensionL of K with [L:K] = n < 03 is graded reaiised if n = [&,(R):R]. L is said to be graded realizable if the gradation of R may be changedas follows, R = Ro[ r, T- ‘1 with deg (T) = t’#O E Z such that [&(R’):Rq = n. Not every finite field extensionL of K is graded realizable over R; examplesand counter-exampleswill be given in Section3.
276 THEOREM 1.9. (Graded version of Zariski’s theorem). Let R be a graded field and let S be a graded field extensionof R generatedas a ring by a definite number of homogeneouselements, xl, . . .,xn say, then S is a graded algebraicextension of R i.e. S is finitely generatedas a gradedR-module.
PROOF. If n = 1 then R[xl] can only be a graded field if xc1 = C rixf with deg (ri) + i deg (xl) = - deg (xl). The manic polynomial deducedfrom (C ?ixi). x1 = 1 has the form (*) and it is easilyseen to be the minimal polynomial of x1 over K, Hence x1 is graded algebraic over R, thus Corollary 1.5. yields that R[xI] is graded algebraicover R and [R[xl]:R] equalsthe degreeof the minimal polynomial for x1. We now proceedby induction, assumingthe statementto be true for graded field extensionsgenerated by lessthan n elements.Proceeding as in the ungradedcase one proves that R[xl, . . ., x,] is an integral extensionof R and then, knowing that R[xI, . . ., x,] is a graded field, it follows immediately that it is a graded algebraicextension of R. H
Making use of this Theorem one proves, exactly as in the ungraded case, the following graded version of Hilbert’s nulstellensatz.
THEOREM 1.lO. If R is a graded field then any proper graded ideal I in S=R[xl, . . . . x,] (graded by S,= C R&LX?) i+v,+...+v,=m has a zero (XI, . . . x,J where the Xilie in a suitablegraded algebraicextension of R.
REMARKS 1.11. 1. A graded field is a Noetherian ring, actually any graded ring which is graded (left) Noetherian is (left) Noetherian as an ungraded ring. 2. If I is a graded ideal of a graded ring R then the intersection of the prime ideals containing I, denoted by rad1, coincideswith the intersection of the graded prime ideals containing I. By the first remark we have that f” E I if and only if fe P for every graded prime ideal P containing I. 3. Note that Theorem 1.10is somewhatdifferent from the well known projective nulstellensatz. Firstly the graded field R is not necessarilytrivially graded. Sincethe extra variable hidden in R i.e. R = Ro[ r, T- I] appearstogether with its inverse, Proj (R[X]) may be considered as an open subspace of Proj(Ro[T,X]), namely the hyperplane “TfO”.
2. INTEGRALEXTENSIONSOFGRADEDRINGS Let R be a graded ring which is a subring of a commutative domain S (not necessarilygraded).
DEFINITION 2.1. An elements of S is said to be graded algebraic over R of
277 degreec, if R[s] sR[X]/Ker Q, whereR[X] is graded by R[X], = Ci+jc=n RjXj and where zs is the ring homomorphism R[X] +S given by 71s(X)= s; z&r) = r for TER, i.e. Ker 7rsis generatedby homogeneouspolynomials of R[X]. We say that s is graded integral over R of degree c, ifs is graded algebraic of degreec and Ker zs contains a manic polynomial.
REMARK 2.2. The definition given is different from the statement:s satisfies a homogeneousmanic polynomial in R[X]. Indeed, evenin the casewhere R is a graded field the “right” definition should imply that the minimal (manic) poIynomia1is homogeneous.
LEMMA 2.3. If s is graded integral over R then there is a homogeneous manic polynomial in Ker ns.
PROOF. Let Xm + Czi’ aiX’ ai ER, be a manic polynomial in Ker 7rs.If X is given degreec, selectingthe homogeneouscomponent of degreemc yields a homogeneousmanic polynomial in Ker fcs. n
LEMMA 2.4. Let SES be such that R[s] (constructed within S) is a graded ring such that s is homogeneousof degreec. If t E S is graded integral over R of degreed, then fis graded integral over R[s] of degreed.
PROOF. By our hypotheseszt:R[X]-)S has graded kernel Ker nr (note that R[X) is now graded such that X has degreed!) and it contains a manic homogeneous polynomial. We have to establish similar properties for TCI:R[S][X]+S, given by n,(A) = A for A ER[s], nt(X) = t. Let h(X) be the manic polynomial in Ker nr. Then also h(X) E Ker nt; we now proceedto show that the homogeneouscomponents of any f(X) E Ker m will again be in Ker m, this being the case,h(X) E Ker zt will finish the proof. First let us point out that if R’ is any domain, f(X), g(X) ER[X] such that degx f(X) ~degx g(X), then there exists ac R’ such that ag(X) = =f(X)q(X) + r(X) with q(X) E R’[X] and degx r(X) c degxf(X).
PART 1. Take g(X) E Ker 7ttrsuch that g(X) has minimal degreein X as such. Taking the homogeneouscomponent of highest degree appearing, will be denoted by putting an index ( -)H while taking components of lowest degree will be denoted by ( -)M.
CASE 1. g(X),vE(R[s][X])o. If alsog(X)ME (R[s][X])o then g(X) is homogeneousand then there isnothing to prove. So let us assumethe statement to be true if g(X)ME (R[s][XJ)f for 0 > l> k. Next consider the casewhere g(X)M E (R[s] [Xl)&. Let a E R[s] be such that ah(X) = g(X)q(X) + r(X). Then ah(t) = 0 and g(t) = 0 yield r(t) = 0 but then degx r(X) c degx g(X) entails r(X) = 0 and ah(X) = g(X)q(X). Since h(X) is homogeneous this yields a&(X) =g~(X)qM(x) thus again gM(t) = 0 or qM(t) = 0. Supposethat qM(f) = 0. Then degx q&X) 1 degx g(X) L degx gM(X). Therefore there is an (X’ER[s] such that &q&X) = g(X)q’(X), hence ohq&X) = gM(X)qh(X). If gM(X) $ R[s] then we may repeat the procedure and since now degx qMX)
CASE 2. g(X)HE (R[s][X])i with iz0. If i= 0 then we are in Case 1, so we proceedby induction on i. ReplacingM by H in the proof for Case 1 yields a completely similar proof.
PART 2. Now considerp(X) E Ker xt of arbitrary degreein X. Let pi(X) be the sum of the homogeneous components of p(X) which are in Ker zt. If p(X) --pi(X) =0 then we are done, otherwise q(X) =p(X) -PI(X) is in Ker zr and no component in q(X) is in Ker ~lt. Let g(X) be as in Part 1, then degx q(X) kdegx g(X), hence we may select y E R[s] such that yq(X) = =g(X)qdX) + r(X) with degx r(X) COROLLARY 2.5. Let R be a graded ring which is a subring of the domain S. Put I(&= (xES, x graded integral of degree c over R} and write I(R) = = @&zl(&. Then I(R) is a graded ring, which is called the graded integral closure of R in S. Note also that in case S is a graded ring too, an element s of S,, which is graded integral over R has to be graded integral of degreen; this showsthat the gradation on I(R) is induced by the gradation of S, in this case. 3. GRADEDVALUATIONRINGSINGRADEDFIELDS Let R be a graded field. A graded subring V of R is said to be a graded valuation ring if for every homogeneousx E R either x or x- l is in V. LEMMA 3.1. If V is a graded valuation ring in R then the graded idealsof V are linearly ordered by inclusion. 279 PROOF. Straightforward. 1 LEMMA 3.2. If al, . . . . a, are homogeneouselements of R-V then there is an i such that a[T’ajE Vfor allj= 1, . . . n. PROOF. Consider the idealsAi generatedby a,: ’ in V. By Lemma 3.1. we may selectan index i such that Ai is minimal amongst the Aj, j= 1 . . . n. Then a,: ‘aj E V. Otherwise aj ’ = (aj ‘ai)a,r ’ yields Aj =Ai but then ra,:’ =alT1 for somerE V i.e. a,: ‘aj E I/ follows again. n With the samenotations: REMARK 3.3. We have that a,: ‘aj EMv (the unique maximal graded ideal of V) if and only if Ai 5 Aj. PROPOSITION 3.4. If V is a graded valuation ring of R with maximal graded ideal Mv then (Mv, V) is a semi-restrictedprime of R (in the senseof [7]). PROOF. Let a=al+ . . . + an ER - V, where ai denotes the homogeneous component of degreem;, Lyle< 1112. , . < m,. Supposeai,, . . . . ai, are in R-V. By Lemma 3.2. we may select b such that alT1ai,E V for ikts {in, . . . . im). Then a~;la=aj;lal+...+l+...+ai;lan~V. If a,: ‘a EMV then, taking parts of degree0, we find 1 + m E (MV)OCMV for some M E (Mv)o, hence1 EMV contradiction. Therefore a,; ‘a E V-MV and, from [7] it results then that (Mv, V) is indeeda semi-restrictedprime in R. Consequently R-V is a multiplicatively closedset! n NOTATION. If S is a subring of a graded ring R then S, will denotethe ring generatedby the homogeneouss E S. If S is an ideal of R then S, is a graded ideal of R; if P is a prime ideal of R then Pg is a graded prime ideal of R. PROPOSITION 3.5. If (P, Rp) is a prime in the graded field R then (Pg, Rf) is also a prime in R and the latter will even be a graded valuation ring with maxi- mal graded ideal Pg. PROOF. Take x homogeneousin R. Sincex x-l E RP it follows that either XE RP or x- 1E RP. Thereforex E Rf or x- 1e R:, henceRf is a graded valuation ring. Furthermore, Pg obviously contains the ideal generatedby the homo- geneousXER such that x-l @Rf, whence it follows that Pg is the maximal graded ideal of Rc. m REMARKS 3.6. 1. Graded valuation rings in graded fields are also restricted primes in the sense of [7]. 2. It is not necessarilytrue that Pf@=P,. Indeed, if this were the casethen (P, Rp) would dominate (Pg, RF) and the latter is a semirestrictedprime thus 280 (cf. [7]) it is maximal with respect to domination i.e. RF= RP and Pg= P follow! A possibleapplication of primes is to the study of integrality and integral extensions,Recall from [7] that an extensionof domains SC r, say, the integral closure of S in T is exactly the intersection of the domains of primes in T containing S. THEOREM 3.7. Let S be a graded domain, let T be a graded field extension over the graded field of fractions of S, then the integral closure of S in T is a graded ring; consequently the integral closure of S in 7’ equals the graded integral closure of S in T. PROOF. The integral closure S of S in T is given by flR’, the intersection ranging over the domains of primes of T containing S. SinceS is graded, this intersection equals II Ri, therefore S is graded. n COROLLARY 3.8. (Theorem 11, [ 141). The integral closure of a graded domain in its field of fractions is a graded ring. THEOREM 3.8. Let R be a graded field with field of fractions K and let L be a finite field extensionof K, [L:K] = n say. Then the following statementsare equivalent: 1. L is graded realizable over R. 2. The integral closure of R in L is a graded ring, containing R as a graded subring if the degreeof T is chosenwell. PROOF. Put R=k[T, T-l], K=k(T). 1 * 2. By 1 we know that R may be graded by chasingdeg (T) = t such that [EL(R):R] = n (note that sinceR is a graded field we may write EL(R) for IL(R)). Consequently,the field of fractions of EL(R) is L. By Theorem 3.7. the integral closure EL(R) of EL(R) in L is graded, but &(A) being a graded field it has to be integrally closedin its field of fractions. Thus a =&(R) is a graded ring. 2* 1. If R is a graded ring containing R as a graded subring (for the gradation on R where this happens)then a homogeneousXE R of degree c satisfies a homogeneous polynomial in R[X] (graded as follows: RIXln= = C;+jc=nRiX)j w h ich may be assumedto be manic sinceR is a graded field. If we derive Sucha manic homogeneouspolynomial from an integral polynomial over R having minimal degree as such then this will also be the minimal polynomial over K for x i.e. XEEL(R) and X has degree c as an element of h(R). n COROLLARY 3.9. The above theorem yields that Theorem 3.7. may be extended to graded realizable extension fields of K. In particular if L =F(t) whereF is a finite extensionof the field k = &, then the integral closure of R in L is a graded ring; this particular case is the generalization of Theorem 11 mentioned in [ 141. 281 EXAMPLES 3. lo. Put R=k[T, FiJ with deg (T)=f;K=k(T). 1. Consider L = K(a) with a2 = T. It is easily seenthat L is graded realizable, indeedif T is given even degree then a E Et(R). However if t is odd then EL(R) = R is easily checked. 2. ConsiderL =K(a) with a2 = T+ 1. Supposewe can give T degreet’ such that L is graded realized over R’. Since ~E:R then a~,?$,@“) by Theorem 3.8. Write then a=ai,+...+ai,, il<... REMARK 3.11. The graded valuation rings of a fixed graded field R form a dense set of Prim R, considered in [ll], with the induced Zariski topology. Thus this is a compact Riemann surface V associatedto the graded field. Note that a copy of the Riemann surface of RO may be found in “Yby consideringthe graded valuation rings OO[T, T- ‘1 where 00 is a valuation ring of Ro. 282 4. GRADEDSPLITTINGFIELDSOFGRADEDSIMPLEARTINIANALGEBRAS THEOREM 4.1.Let D be a graded division ring. Then either D=& is a skewfield or elseD =Do[X, X-l, (p], where DOis a skewfield and c~is an auto- morphism of DO, where X is a variable with deg (X) >O. PROOF. Cf. [5]. n For more details on graded simple graded Artinian rings we refer to [5]. Let us point out here that these graded definitions are indeed the obvious graded versions of the ungraded definitions. Recall also that the graded version of Wedderburn’s theorem tells us that everygraded simple graded Artinian ring is isomorphic to M@)(d) for some graded division ring D. Recall that the gradation on &(D)(d) with (d)~ Zn is as follows: Di+dl-62 . . . Di+Dl -dn DiDi+dz-dl Di i (Mn(D)W)i = : : : : Di+dn-dl 0 > We look at graded simple Artinian rings which are uf finite rank over their center. So if D is trivially graded then D = Do is a central simple algebra; if the gradation is non-trivial then D = Do[X, X- l, (p] where p is an automorphism of Do such that 9” is inner, in that casethe center of D is kp[ T, T- ‘1where kp is the field left fixed by Q in the center of DO and where T= AX” for some J. EDO - (0); consequentlythe rank of D over k,[T, T-I] is then equal to (nq)2 where q2 is the dimension of Do over its center. Finally let us point out that the number of non-equivalent A&(D)(d) for do Z” is finite if D is non-trivially gradedbut may be infinite otherwise! In the sequelwe will considerD which are non-trivially graded. From [ 121it follows then that Do[X, cp]is a Zariski central ring with center k,[T], T= AX”, hence its skewfield of fractions Do(X, p) has k&T) for its center and Do(X, p) has rank (nq)2 over k,(T). THEOREM 4.2. Let D be a graded division ring of finite rank over its center. Let C be a maximal commutative graded subring of D. Then C is a graded field and a maximal commutative subring of D, we have that: D 0 Cs M&C)(d) for somedo k”q. Z(D) PROOF. Obviously C is a graded field. If r E D is such that rc = cr for all c E C then rci = cir for all homogeneousci of C. Put r = ri, + . . . + rj,, r$ being the components of r, then comparing degrees learns that cirG=rbci and thus rbc = crb for every c E C. The assumptionon C yields then rb E C for everyj, i.e. r E C, and this shows that C is a maximal commutative subring of D. 283 The remarks preceding this theorem yield that the field of fractions of D is obtainedby central localization at the prime ideal 0 of kV[T, T- ‘3. Let us denote Q(C) for the ring of fractions &(T)@C, which is a communative subring of Do(X, p). If r E Do(X, p) commutes with Q(C) then sy E Do[X, X- l, cp]for some SEDo[X, X-l, Do[X, X-l, ~3 for someSE k,[T, T- i] = Z(D) and of coursesy commutes with C i.e. syg C or YE Q(C). Now Q(C) being a maximal commu- tative subring of Do(X, q) it has to be a splitting field of Do(X, ~0).Exactness of the central localization functor then yields that: Q REMARK 4.3. The ungradedtheorems referred to in (*) may for examplebe found in [3], the proofs given there may be easily adjusted to the graded case. EXAMPLE 4.4. Consider D = C[X, X- ‘, ~,3 wherep is conjugation in C. The center of D is lR[X2, Xm2] and D has rank 4 over lR[X2, XP2]. It is clear that R[X2,XS2] and R[X, X- t ] are gradedsplitting fields of D. Obviously C(X2) and R(X) are splitting fields of c(X, p) and both are graded realizedfield extensions of R(X2). On the other hand, R[Xe2, X-t i] is a maximal commutative subring of D which is graded non-realizable. Indeed (X+ i)2 =X2 - 1, hence lR(XV2, X+ i) = lR(Xf i) is of the type consideredin Examples3.10 (2). COROLLARY 4.5. 00(X, (p) has graded realizablesplitting fields. PROOF. It is easily checkedthat the maximal graded commutative subrings of Do[X, X-l, cp]yield graded realizablesplitting fields of Do(X, p). W Amongst the graded realizable splitting fields of Do(X, p) there are two specialtypes to be singledout. A graded realizedsplitting field L of&(X, q$ is said to be o-maximal if L = I(T) for some maximal subfield I of DO; L is called o-minimal if L = f’(X) for somesubfield I’ of DO. 284 PROPOSITION 4.6. There exist o-maximal and o-minimal splitting fields for Do(X cp). PROOF. 1. Consider the fixed ring a of ~0within DO, then a is a skewfield with center, kq say. Let I’ be a maximal subfield of D$. (The easiestway to establishthat a is finite dimensionalover its centeris to use the fact that D is a P.I. ring hence so is ~cD, Posner’s theorem and the fact that a is its own ring of quotients then yield that a is finite over its center). If T=ilx” then A E a is easilyverified so we can chooseI+’ such that 1 E I+‘.Consider Ig[X, X- ‘1 and suppose that P(X) EDo[X, X-l, (p] commutes with all elements of Ip[X, X- l]. We may write P(X) = C ‘icza;Xi (convention: we write the elements of Do[X, X- ‘, cp]this way!). Clearly XP(X) - p(X)X= 0 yields a; E DCD.Further, if c E Iv then cp(X) = P(X)c yieldsai = caic- ‘, thus the choiceof Iv impliesaie Iv i.e. P(X) E 1v[X, X-l]. Note that A E Z’J’was necessary in order to have T=AX” E Ip[X, X-t]. We have shown that /p[X, X-l] is a maximal commutative graded subring of D, hence/q(X) is a o-minimal splitting field for Do(X VP). 2, Let I be a maximal subfield of Do, and consider f[ r, T- ‘1, Supposethat P(X) = C’iE~ajXi commutes with I[T, T- ‘1. Take ,uE Z(DO)c I then pP(X) = P(X) yields $01) =p. Hence I$ is the identity in Z(D0) i.e. (p’is inner in DO and from this it follows that i is an n-multiple. This meansthat C’iEzaiXi may be written as C ‘iEhalT”i. For every y E I, VP(X) =P(X)y then yields a,!y= ya,f or a/E I and then P(X) E I[7’, T-l] is evident. Thus I(T) is a o-maximal splitting field of DOW, ~1. n COROLLARY 4.7. (Graded version of Posner’s theorem). A positively graded prime ring S is a P.I. ring if and only if S is a graded subring of M,,(R)(d) for some graded field R and somedo Z”. PROOF. If S+M,@)(d) then S is clearly a P.I. ring, Conversely, if S is a positively graded P.I. ring then it is a graded prime Goldie ring and it has a graded simple graded Artinian ring of quotients Qg(S) which may be obtained from S by inverting the homogeneousregular elementsof S (using the methods of [5] it is possibleto prove directly that this set is indeed a left Ore set). We have that Qg(S)=A& (D)(d’) for some d’ E H”, and some graded skewfield D. Supposethat D = DO. Sincethe center of D is the ring of graded fractions of the centerof S (cf. [ 131)it follows thus that Z(S) is trivially graded. Take xf 0 is S, for somep#O. The ideal SxS intersectsthe center in a non-zero graded ideal, thus there exrst si,, . . .,si, and sil, . . ., s{, such that Es+ X.S~~EZ(S), and the s$ and the sh may be chosento be homogeneous,of degreet+ resp. t; say, SinceS is positively graded t$ +p + tj= 0 is impossiblehence D cannot be trivially graded if S is non-trivially graded. Let Q(S) be the total ring of fractions, then Q(S) = Mn(Q(D)) = M,(Do(X, p)). It follows that Q(D) is finite dimensionalover its centre and as Q(D) = D@~,[T, r- Ilkp(T), D is of finite rank over k,[T, T- I]. 285 So we may choose a maximal graded commutative subring (of Do[X, X- I, q] = D) and then we get: (all morphisms being morphisms of graded rings). REMARK 4.8. 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